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May 5, 2015 at 10:57 history edited zrysky CC BY-SA 3.0
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May 5, 2015 at 10:51 history edited zrysky CC BY-SA 3.0
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May 5, 2015 at 10:00 comment added zrysky @EdenHarder Yes, because we only extend $t$ to complex plane, but $\mathbf{x}$ is always real.
May 5, 2015 at 9:53 comment added Eden Harder I know, it's because $t = \pm (|{\bf x}|+i\epsilon)$ and $i|{\bf x}|^2$ is totally imaginary.
May 5, 2015 at 9:49 comment added Eden Harder Thanks a lot! But why ${\rm Re}(ix^2) = {\rm Re}(it^2)$? Generally, ${\rm Re}(ix^2) = {\rm Re}(it^2-i|{\bf x}|^2)$.
May 5, 2015 at 9:24 history bounty ended Eden Harder
May 5, 2015 at 9:24 vote accept Eden Harder
May 5, 2015 at 8:11 review First posts
May 5, 2015 at 8:24
May 5, 2015 at 8:10 history answered zrysky CC BY-SA 3.0